Quadratic Equations Test

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Quadratic Equations Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Solve $$9m^2-12m+2=0$$ by method of completing square.

A
$$\dfrac {2+\sqrt 2}{3}$$

B
$$\dfrac {\sqrt 2}{3}$$

C
$$2$$

D
$$\dfrac {-2+\sqrt 2}{3}$$

Solution

$$9m^2-12m+2=0$$
$$9m^2-12m=-2$$
Dividing both sides by $$9$$, we get
$$m^2-\dfrac {12}{9} = \dfrac {-2}{9}$$

$$m^2-\dfrac {4}{3} = \dfrac {-2}{9}$$

Third term = $$\left(\dfrac 12 \times \mbox{coefficient of m}\right)^2 $$

$$=\left (\dfrac 12 \times \dfrac {-4}{3}\right)^2 = \dfrac 49$$

Adding $$\dfrac 49$$ on both sides, we get

$$m^2-\dfrac 43 m+ \dfrac 49 = \dfrac {-2}{9} + \dfrac 49$$

$$\left(m-\dfrac 23\right)^2 = \dfrac 29$$

Taking square roots on both sides we get,

$$\left(m-\dfrac 23\right) =\pm \dfrac{\sqrt{2}}3$$

$$m=\dfrac {2+\sqrt 2}{3}$$ or $$m=\dfrac {2-\sqrt 2}{3}$$
Question 2
1 / -0

The roots of the equation $$\displaystyle x^{2}-px+q=0$$ are consecutive integers. Find the discriminate of the equation.

A
$$-1$$

B
$$0$$

C
$$1$$

D
None of these

Solution

For the given equation, let the roots be $$ a, a+1 $$

So sum of roots $$ = a + a + 1 = 2a + 1= -\cfrac {-p}{1} = p $$

Product of roots $$ = a \times (a+1) = {a}^{2} + a= \cfrac {q}{1} = q $$

For an equation $$ a{x}^{2} + bx + c = 0 $$, the discriminant is $$ {b}^{2}
-4ac$$.

For the given eqn, the discriminant is $${ (-p)}^{2} -4\times 1 \times q = (2a + 1)^{2}- 4({a}^{2} + a) = 4{a}^{2} + 1 + 4a - 4{a}^{2} -4a = 1 $$
Question 3
1 / -0

Solve: $$\displaystyle x^{2}+3x+1=0 $$

A
$$\displaystyle x=\frac{3\pm\sqrt{5}}{2}$$

B
$$\displaystyle x=\frac{3+\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}$$

C
$$\displaystyle x=\frac{\pm{3}-\sqrt{5}}{2}$$

D
$$\displaystyle x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$$

Solution

We have $$\displaystyle x^{2}+3x+1=0$$
Add and subtract $$\displaystyle \left ( \frac{1}{2}\mbox{coefficient of x} \right )^{2}$$ in LHS and get $$\displaystyle x^{2}+3x+1+\left ( \frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2}=0 $$

$$\displaystyle \Rightarrow x^{2}+2\left ( \frac{3}{2} \right )x+\left ( \frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2}+1=0 $$

$$\displaystyle \Rightarrow \left ( x+\frac{3}{2} \right )^{2}-\frac{5}{4}=0 $$

$$\displaystyle \Rightarrow \left ( x+\frac{3}{2} \right )^{2}=\left ( \frac{\sqrt{5}}{2} \right )^{2}$$

$$\displaystyle \Rightarrow x+\frac{3}{2}=\pm \frac{\sqrt{5}}{2}$$

This gives $$\displaystyle x=\frac{-3+\sqrt{5} }{2}$$ or $$x=\dfrac{-3-\sqrt{5}}{2}$$

Therefore, $$\displaystyle x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$$ are the solutions of the given equation
Question 4
1 / -0

If $$\displaystyle l^{2}+m^{2}+n^{2}=5,$$ then (Im+mn+In) is

A
$$\displaystyle \geq (-5/2)$$

B
$$\displaystyle \geq (-1)$$

C
$$\displaystyle \leq 5$$

D
$$\displaystyle \leq 3$$

Solution

We know that

$$ {(l+m+n)}^{2} = {l}^{2} + {m}^{2} + {n}^{2} + 2(lm+mn+ln) $$
$$ => {(l+m+n)}^{2} = 5 + 2(lm+mn+ln) $$

We know that $$ {(l+m+n)}^{2} $$ will be a value greater than or equal to zero

$$ => {(l+m+n)}^{2} \ge 0 $$
$$ => 5 + 2(lm+mn+ln) \ge 0 $$
$$ => (lm+mn+ln) \ge =\frac {5}{2} $$
Question 5
1 / -0

Find the discriminant of $$\displaystyle { x }^{ 2 }-5x-10=0$$

A
$${65}$$

B
$${-65}$$

C
$$10$$

D
None of the above

Solution

Discriminant is given as:
$$\displaystyle D={ b }^{ 2 }-4ac$$
$$\implies D={ \left( -5 \right) }^{ 2 }-4\left( -10 \right) $$
$$\displaystyle =25+40$$
$$\displaystyle =65$$
Question 6
1 / -0

Which of the following is not a quadratic equation

A
$$\displaystyle x-\frac { 3 }{ 2x } =5$$

B
$$\displaystyle 4x-\frac { 5 }{ 8 } ={ x }^{ 2 }$$

C
$$\displaystyle x+\frac { 1 }{ x } =9$$

D
$$\displaystyle 4x-\frac { 2 }{ 3x } =4{ x }^{ 2 }$$

Solution

Lets, simplify all the equations given in options and convert them to standard forms,

(A) $$\displaystyle x-\frac { 3 }{ 2x } =5$$
Now,
$$\displaystyle x-\frac { 3 }{ 2x } =5$$

$$\Rightarrow \displaystyle \frac {2x^2 - 3 }{ 2x } =5$$

$$\Rightarrow \displaystyle 2x^2 - 3 =10x$$

$$\Rightarrow \displaystyle 2x^2 -10x- 3 =0$$

The highest power of variable $$x$$ is $$2$$. So, it is an quadratic equation.

(B) $$\displaystyle 4x-\frac { 5 }{ 8 } =x^2$$
Now,
$$\displaystyle 4x-\frac { 5 }{ 8 } =x^2$$

$$\Rightarrow \displaystyle \frac {32x - 5 }{ 8 } =x^2$$

$$\Rightarrow \displaystyle 32x - 5 =8x^2$$

$$\Rightarrow \displaystyle 8x^2 -32x +5 =0$$

The highest power of variable $$x$$ is $$2$$. So, it is a quadratic equation.

(C) $$\displaystyle x+\frac { 1 }{ x } =9$$
Now,
$$\displaystyle x+\frac { 1 }{ x } =9$$

$$\Rightarrow \displaystyle \frac {x^2 + 1 }{ x } =9$$

$$\Rightarrow \displaystyle x^2 + 1 =9x$$

$$\Rightarrow \displaystyle x^2 -9x+ 1 =0$$

The highest power of variable $$x$$ is $$2$$. So, it is a quadratic equation.

(D) $$\displaystyle 4x-\frac { 2 }{ 3x } =4x^2$$
Now,
$$\displaystyle 4x-\frac { 2 }{ 3x } =4x^2$$

$$\Rightarrow \displaystyle \frac {12x^2 - 2 }{ 3x } =4x^2$$

$$\Rightarrow \displaystyle 12x^2 - 2 =12x^3$$

$$\Rightarrow \displaystyle 12x^3 -12x^2+2 =0$$

$$\Rightarrow \displaystyle 6x^3 -6x^2+1 =0$$

Here, the highest power of variable $$x$$ is $$3$$. So, it is not a quadratic equation.

Hence, $$Op-D$$ is correct.
Question 7
1 / -0

When solved using the quadratic formula, the solutions to the equation $$3x^{2} - 4x - 6 = 0$$, rounded to the nearest hundredth, are

A
$$2.46, -3.79$$

B
$$2.23,-0.90$$

C
$$3.79,-2.46$$

D
$$0.90,-2.23$$

Solution

$$\Rightarrow$$ Given quadratic equation is $$3x^2-4x-6=0$$, comparing it with $$ax^2+bx+c=0$$
$$\Rightarrow$$ We get, $$a=3,\,b=-4,\,c=-6$$
$$\Rightarrow$$ $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(3)(-6)}}{2(3)}$$

$$=\dfrac{4\pm\sqrt{16+72}}{6}$$

$$=\dfrac{4\pm\sqrt{88}}{6}$$

$$=\dfrac{4\pm2\sqrt{22}}{6}$$

$$=\dfrac{2\pm\sqrt{22}}{3}$$

$$=\dfrac{2\pm4.69}{3}$$

$$\Rightarrow$$ $$x=\dfrac{2+4.69}{3}$$ and $$x=\dfrac{2-4.69}{3}$$

$$\therefore$$ $$x=2.23$$ and $$x=-0.90$$
Question 8
1 / -0

Solve $$x^2+8x+15=0$$ by method of completing square.

A
$$x=3, -5$$

B
$$x=-3, -5$$

C
$$x=3, 5$$

D
$$x=-3, 5$$

Solution

Given:
$$x^{2}+8x+15=0$$
$$x^{2}+8x=-15$$
Add $$16$$ on both sides,
$$x^{2}+8x+16=-15+16$$
$$x^{2}+2\times4x+4^{2}=1$$
$$(x+4)^{2}=1$$
$$x+4=\pm1$$
Hence, $$x=-4+1=-3$$ or $$x=-4-1=-5$$
Question 9
1 / -0

Given that $${ z }^{ 2 }-10z+25=9$$, what is $$z$$?

A
$$ {3,4} $$

B
$$ {1,6} $$

C
$$ {2,6} $$

D
$$ {2,8} $$

Solution

Since you recognize that the left hand side of the equation is a perfect square quadratic, you will factor the left side of the equation first, instead of trying to set everything equal to $$0$$
$${ z }^{ 2 }-10z+25=9$$
$${(z-5)}^{2}=9$$
$$\implies z-5=\pm 3$$
$$\implies z=5\pm 3$$
$$\therefore z= 2$$ or $$z=8$$
Question 10
1 / -0

If $$\displaystyle xy\neq 0$$ and $$\displaystyle { x }^{ 2 }{ y }^{ 2 }-xy=6$$, which of the following could be y in terms of x ?
I. $$\displaystyle \frac { 1 }{ 2x } $$
II. $$\displaystyle -\frac { 2 }{ x } $$
III. $$\displaystyle \frac { 3 }{ x } $$

A
I only

B
II only

C
I and II only

D
I and III only

E
II and III

Solution

Solving the quadratic in y
$$ x^2y^2 -xy - 6 = 0$$
$$\rightarrow$$ y = $$\cfrac{x +/- \sqrt {x^2 + 24x^2}}{2x^2}$$
$$\rightarrow$$ y = ($$x$$ +/- $$5x$$)/(2$$x^2$$)
y = $$\cfrac{3}{x}$$ or y = $$\cfrac{-2}{x}$$
Option E.

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Quadratic Equations Test - 31

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Quadratic Equations Test - 31

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SHARING IS CARING

Question 1

$$\dfrac {2+\sqrt 2}{3}$$

$$\dfrac {\sqrt 2}{3}$$

$$2$$

$$\dfrac {-2+\sqrt 2}{3}$$

Solution

Question 2

$$-1$$

$$0$$

$$1$$

None of these

Solution

Question 3

$$\displaystyle x=\frac{3\pm\sqrt{5}}{2}$$

$$\displaystyle x=\frac{3+\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}$$

$$\displaystyle x=\frac{\pm{3}-\sqrt{5}}{2}$$

$$\displaystyle x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$$

Solution

Question 4

$$\displaystyle \geq (-5/2)$$

$$\displaystyle \geq (-1)$$

$$\displaystyle \leq 5$$

$$\displaystyle \leq 3$$

Solution

Question 5

$${65}$$

$${-65}$$

$$10$$

None of the above

Solution

Question 6

$$\displaystyle x-\frac { 3 }{ 2x } =5$$

$$\displaystyle 4x-\frac { 5 }{ 8 } ={ x }^{ 2 }$$

$$\displaystyle x+\frac { 1 }{ x } =9$$

$$\displaystyle 4x-\frac { 2 }{ 3x } =4{ x }^{ 2 }$$

Solution

Question 7

$$2.46, -3.79$$

$$2.23,-0.90$$

$$3.79,-2.46$$

$$0.90,-2.23$$

Solution

Question 8

$$x=3, -5$$

$$x=-3, -5$$

$$x=3, 5$$

$$x=-3, 5$$

Solution

Question 9

$$ {3,4} $$

$$ {1,6} $$

$$ {2,6} $$

$$ {2,8} $$

Solution

Question 10

I only

II only

I and II only

I and III only

II and III

Solution

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